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MA 3502 Regression Analysis and Experimental Design
Structure of Examination Paper:
There are Five pages.
There are FOUR questions in total.
There are no appendices.
The mark obtainable for each question is 25.
Answer FOUR questions.
Show clearly all your workings.
In all exercises, dj is the j-th digit of your student number.
1a. Consider the linear regression model Y = Xθ+ε, where θ is an m-vector of unknown parameters, Y is an N-vector of observations, X is a design matrix of size N × m and ε is a vector of normal random errors with mean 0 and covariance matrix σ2IN . Here σ2 is some positive unknown constant and IN is the identity matrix of size N × N. Assume that rank(X) = p ≤ m.
(i) Give a geometrical interpretation of the LSE, the least squares estimator (LSE) of θ.
(ii) Defifine the lasso estimators of θ and discuss a motivation behind these estimators.
(iii) Defifine the ridge estimator of θ and brieflfly discuss its properties.
(iv) Let Z be a vector of size m such that the linear form ZT θ is estimable. Explain the method of constructing confifidence intervals for ZT θ.
(v) Derive the maximum likelihood estimator for σ2.
[10=2+2+2+2+2]
1b. Assume m = 3, N = 5 and the model yj = θ0 + θ1xj + θ2x2j + εj . The table below displays the design points and the corresponding results:
xj 0 u −u v −v
yj 0 1 −1 1 −1
Here u = d4 + 1 and v = d4 + 2, where d4 is the fourth digit of your student number.
(i) Test the hypothesis that the complete regression model is statistically signifificant on the 95% confifidence level. Write down the corresponding ANOVA table and compute R2.
(ii) Test the hypothesis
on the 95% confifidence level.
To answer this question please use 19.0 as an approximate value for the 0.95-quantile of the F-distribution with 2 and 2 degrees of freedom.
[15=7+8]
2a. Consider the linear regression model Y = Xθ+ε, where θ is an m-vector of unknown parameters, Y is an N-vector of observations, X is a design matrix of size N × m and ε is a vector of normal random errors with mean 0 and covariance matrix σ2IN .
In a particular application, m = 4, N = 5 and the model is Y = θ1X1 + θ2X2 + θ3X3 + θ4X4 + ε. The table below displays the design points and the corresponding results
X1 u -1 1 0 -1
X2 −v 1 0 1 -1
X3 u−v 0 1 1 -2
X4 u+2v -3 1 -2 1
Y 1 1 1 -1 0